- #36
- 10,302
- 1,472
It's messy, but my computer algebra package is confirming the solution.
The highlights:
$$g(r) = \frac{df(r)}{dr} \quad \dot{t}(\tau)= \frac{E}{f(r(\tau))} \quad \dot{r}(\tau) = \sqrt{E^2 - f(r(\tau))} $$
$$ \ddot{t}(\tau) = -\frac{E\,g(r)}{f^2(r)} \dot{r}= \frac{-E\, g(r(\tau)) \sqrt{ E^2 - f(r(\tau))} } {f^2(r(\tau))} $$
$$ \ddot{r}(\tau) = -\frac{1}{2} \, \frac{g(r)} { \sqrt{E^2-f(r)}} \dot{r} = -\frac{g(r(\tau)) } {2 }$$
The rest is more algebra, substuting ##\ddot{t}## ##\dot{t}## ##\ddot{r}## and ##\dot{r}## into the geodesic equations.
Note that when g(r)=0, ##\ddot{r}=0##, which is consistent with the idea that f(r) is the effective potential, and the metastable peak of the effective potential occurs when g(r) = df/dr = 0.
From the expression for ##\dot{r}## we could write
$$\frac{dr}{\sqrt{E^2 - f(r)}} = d\tau$$
which could be integrated to solve for ##\tau(r)## which could be inverted to get ##r(\tau)##, but this gets very messy very quickly.
The highlights:
$$g(r) = \frac{df(r)}{dr} \quad \dot{t}(\tau)= \frac{E}{f(r(\tau))} \quad \dot{r}(\tau) = \sqrt{E^2 - f(r(\tau))} $$
$$ \ddot{t}(\tau) = -\frac{E\,g(r)}{f^2(r)} \dot{r}= \frac{-E\, g(r(\tau)) \sqrt{ E^2 - f(r(\tau))} } {f^2(r(\tau))} $$
$$ \ddot{r}(\tau) = -\frac{1}{2} \, \frac{g(r)} { \sqrt{E^2-f(r)}} \dot{r} = -\frac{g(r(\tau)) } {2 }$$
The rest is more algebra, substuting ##\ddot{t}## ##\dot{t}## ##\ddot{r}## and ##\dot{r}## into the geodesic equations.
Note that when g(r)=0, ##\ddot{r}=0##, which is consistent with the idea that f(r) is the effective potential, and the metastable peak of the effective potential occurs when g(r) = df/dr = 0.
From the expression for ##\dot{r}## we could write
$$\frac{dr}{\sqrt{E^2 - f(r)}} = d\tau$$
which could be integrated to solve for ##\tau(r)## which could be inverted to get ##r(\tau)##, but this gets very messy very quickly.